1046. Shortest Distance (20)

1046. Shortest Distance (20)

時間限制

100 ms

內存限制

65536 kB

代碼長度限制

16000 B

判題程序

Standard

作者

CHEN, Yue

The task is really simple: given N exits on a highway which forms a simple cycle, you are supposed to tell the shortest distance between any pair of exits.

Input Specification:

Each input file contains one test case. For each case, the first line contains an integer N (in [3, 10⁵]), followed by N integer distances D₁ D₂ ... D_N, where D_i is the distance between the i-th and the (i+1)-st exits, and D_N is between the N-th and the 1st exits. All the numbers in a line are separated by a space. The second line gives a positive integer M (<=10⁴), with M lines follow, each contains a pair of exit numbers, provided that the exits are numbered from 1 to N. It is guaranteed that the total round trip distance is no more than 10⁷.

Output Specification:

For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.

Sample Input:

5 1 2 4 14 9
3
1 3
2 5
4 1

Sample Output:

3
10
7

本題由於圖形簡單，任意兩點之間只有兩條路可以選擇，而且所有的點又組成了環，這兩條路徑的和總是等於環的周長，當輸入完畢時，我們就可以知道整個環的“周長”，同時在輸入時記錄每個點到起點（第一個點）的距離，測試時，用兩點到起點的距離相減就是兩點之間的距離dis，同時，用周長減去dis就是第二條路徑的長度，輸出兩者中較小者即可。

#include <cstdio>

using namespace std;
#define BIG 100001

int min(int a,int b){
	if(a >= b)
		return b;
	else 
		return a;
}

int main(){
	int sum = 0, toFirst[BIG];
	int m,n,a,b,temp = 0,to;
	scanf("%d",&n);

	toFirst[1] = 0;
	for (int i = 2; i <= n; ++i)
	{
		scanf("%d",&temp);
		toFirst[i] = toFirst[i-1] + temp;//每個點到第一個點的距離
		sum += temp;
	}
	scanf("%d",&temp);
	sum += temp;//周長

	scanf("%d",&m);
	for(int i=1;i<=m;i++){
		scanf("%d %d",&a,&b);
		if(a > b){
			a ^= b ^= a ^= b;//交換a和b的值
		}
		printf("%d\n", min(toFirst[b]-toFirst[a], sum-(toFirst[b]-toFirst[a]) ) );
	}

	return 0;
}